Answer:
3 > x
Step-by-step explanation:
0>3x-3-6
Combine like terms
0>3x-9
Add 9 to each side
0+9>3x-9+9
9 > 3x
Divide each side by 3
9/3 > 3x/3
3 > x
The answer is less than or equal to -4. Because -4+8=4
1) True that line x = 0 is perpedicular to y = -3. Because x = 0 is parallel to the y-axis and y = -3 is parallel to the x-axis.
2) True that all the lines that are parallel to the y-axis are vertucal lines (the y-axis is vertical)
3) False that all lines perpendicular to the x-axis have a slope o 0. Their slope trends to infinity.
4) False that the equation of the line parallel to the x-axis that passes through the point (2, –6) is x = 2. The right equation is y = - 6
5) True thath the equation of the line perpendicular to the y-axis that passes through the point (–5, 1) is y = 1
Answer:
Let's define the cost of the cheaper game as X, and the cost of the pricer game as Y.
The total cost of both games is:
X + Y
We know that both games cost just above AED 80
Then:
X + Y > AED 80
From this, we want to prove that at least one of the games costed more than AED 40.
Now let's play with the possible prices of X, there are two possible cases:
X is larger than AED 40
X is equal to or smaller than AED 40.
If X is more than AED 40, then we have a game that costed more than AED 40.
If X is less than or equal to AED 40, then:
X ≥ AED 40
Now let's take the maximum value of X in this scenario, this is:
X = AED 40
Replacing this in the first inequality, we get:
X + Y > AED 80
Replacing the value of X we get:
AED 40 + Y > AED 80
Y > AED 80 - AED 40
Y > AED 40
So when X is equal or smaller than AED 40, the value of Y is larger than AED 40.
So we proven that in all the possible cases, at least one of the two games costs more than AED 40.
Answer:
7
Step-by-step explanation: